Steve said:
>Yes, and maybe I should have said 'higher' rather than just 'high' - but
are amplifiers unstable if you disconnect the blocking cap and introduce
infinte R?<
As usual - it depends! Because there's inherent inductance and shunt
capacitance, there's obviously an impedance at VHF. The higher the impedance,
the higher the gain. Stability depends on the phase and magnitude of feedback.
An infinite anode load means that gain will become gm divided by Yload, where Y
load is the sq. rt of R squared plus X squared. R is the load resistance in
shunt with the ra of the valve. [a variant on 'muRL/(ra+RL)]. So the gain will
have a maximum value that's likely to be higher than that achieved with a shunt
L-R network. Whether or not that leads to oscillation then depends on the Bode
plot parameters for the feedback.
I seem to remember that the shunt network should only get the resisitors hot
from fundamental frequency current, because there isn't any parasitic current
if the the thing hasn't got parasitics. 100nH at 28MHz is about 17 ohms, so
with about 1 amp rms effective anode current flowing, that suggests a shunt 47
ohm resistor needs to be able to disipate 6.5watts. As it's in a hot
environment, this suggests that a 'Globar' type resistor rather than a couple
of plain old carbon comps is required. That is one advantage of the nichrome
wire - it doesn't go high resistance with time - well not unless you're getting
it red hot repeatedly!
73
Peter G3RZP
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